How would you prove mathematically that electric field inside a conductor is zero? I know the arguments but I am looking for a mathematical proof for the same.

  • $\begingroup$ In a conductor, the electric field and current density are linearly related $\vec {j}=\sigma \vec {E}$. In electrostatics $\vec {j}=0$, therefore $\vec {E}=0$. $\endgroup$ – Alex Trounev Jan 28 '20 at 13:11
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    $\begingroup$ That is an answer, not a comment. $\endgroup$ – my2cts Jan 28 '20 at 13:29
  • $\begingroup$ Does this answer your question? Look for the part on conductors. $\endgroup$ – aditya_stack Jan 28 '20 at 16:26
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    $\begingroup$ Does this answer your question? How to show mathematically that the electric field inside a conductor is zero? $\endgroup$ – sammy gerbil Jan 29 '20 at 20:27
  • $\begingroup$ When you post a new question you are given suggestions which might already give the answer you are looking for. Please look at them. If one of the suggestions is asking the same question as you, then you need to explain why the answers to that question are not satisfactory for you. $\endgroup$ – sammy gerbil Jan 29 '20 at 20:33

The simplest proof is that for a conductor the current density inside the conductor is given by:

$$ \mathbf J = \sigma \mathbf E $$

where $\sigma$ is the conductivity and $\mathbf E$ is the field inside the conductor. At equilibrium the current density has to be zero everywhere inside the conductor, and this is only possible if $\mathbf E$ is zero everywhere inside the conductor.

A corollary of this is that all the charge has to reside on the surface of the conductor. If the field is zero inside the conductor then the flux through any closed Gaussian surface inside the conductor is zero, and that means the charge enclosed by that surface must be zero.

  • $\begingroup$ We apply the approximation of ${\bf E}=0$ in many situations that aren't strictly electrostatics. In DC steady-state circuits. Even in RF electromagnetics. $\endgroup$ – The Photon Jan 28 '20 at 16:54

You can simply prove it by gauss law_

Take a closed surface ( inside the solid conductor) , if there is no charge within this , there will be no flux.hence no electric field.

The matter of the fact is that charge only exist on the outer surface of the conductor.( I am taking a solid conductor with no cavity)

If it would not be a conductor , then charge will somehow distributed within the material , and any closed surface has some charge within it .therefore, the field will not be zero. One always has to to assume one of the properties of conductor to prove this so I have taken the fact that charge always resides on the surface. therefore from the Gauss law there can't be any electric field inside the solid conductor

  • $\begingroup$ I think this is a circular argument. You are starting with the assumption that the charge is all on the surface, but the charge is all on the surface because the field inside has to be zero. $\endgroup$ – John Rennie Jan 28 '20 at 15:23
  • $\begingroup$ I am proving this by the fact that charge always resides on the surface . $\endgroup$ – Kunal kumar Jan 28 '20 at 15:26
  • $\begingroup$ You have to take one of the property of the conductor to prove this mathematically. $\endgroup$ – Kunal kumar Jan 28 '20 at 15:27

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